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Homework Help: Potential Energy of a system of Point Charges

  1. Aug 5, 2007 #1
    1. The problem statement, all variables and given/known data

    Two identical charges q are placed on the x axis, one at the origin and the other at x = 5 cm. A third charge -q is placed on the x axis so the potential energy of the three-charge system is the same as the potential energy at infinite separation. Its x coordinate is: (x=13)

    2. Relevant equations

    I know the correct answer to this one because I'm looking over a marked up test in preparation for the final.

    I've been using the equations for the electric potential energy of a system of point charges:

    [tex]U=W=k\frac{q_{1}q_{2}}{r}[/tex]

    3. The attempt at a solution

    ...and expanding that to the work of the entire system:

    [tex]W=0=W_{12}+W_{23}+W_{13}[/tex]

    so,


    [tex]W=0=k\frac{q_{1}q_{2}}{5}+k\frac{q_{2}q_{3}}{d}+k\frac{q_{1}q_{3}}{5+d}[/tex]

    Since [tex]q_{3}[/tex] is negative, simplify:

    [tex]W=0=k\frac{q^{2}}{5}-k\frac{q^{2}}{d}-k\frac{q^{2}}{5+d}[/tex]


    I might be missing something simple in algebra, but I can't figure out how to simplify this down enough to solve for [tex]d[/tex].

    Any ideas? Am I even on the right track here?

    Thanks!
     
  2. jcsd
  3. Aug 6, 2007 #2
    [tex]kq^2[/tex] can be eliminated (as the right hand side is equal to zero), and this leaves you with:

    [tex]\frac{1}{5} -\frac{1}{d} -\frac{1}{5+d}=0[/tex]. Take the negative terms to one side and cross multiply to get a quadratic equation in d. Solve quadratic for d. There are two possible solutions as this is a quadratic.
     
  4. Aug 6, 2007 #3
    I'm sorry, I forgot to mention that I tried doing a quadratic, but was unable to get the correct answer. Also, I think that the correct term should be [tex]\frac{1}{(d-5)}[/tex], not [tex]\frac{1}{5+d}[/tex].
     
  5. Aug 7, 2007 #4
    Yes. You're right. I didnt look at the whole post. The solution, however, is [tex]x=\frac{15}{2}+-\frac{5\sqrt{5}}{2}[/tex]
     
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